\(\int \frac {(1+x^8)^{3/4}}{-1+x^8} \, dx\) [2099]

Optimal result

Integrand size = 17, antiderivative size = 152 \[ \int \frac {\left (1+x^8\right )^{3/4}}{-1+x^8} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^8}}\right )}{4 \sqrt [4]{2}}+\frac {\arctan \left (\frac {2^{3/4} x \sqrt [4]{1+x^8}}{\sqrt {2} x^2-\sqrt {1+x^8}}\right )}{4\ 2^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^8}}\right )}{4 \sqrt [4]{2}}-\frac {\text {arctanh}\left (\frac {2 \sqrt [4]{2} x \sqrt [4]{1+x^8}}{2 x^2+\sqrt {2} \sqrt {1+x^8}}\right )}{4\ 2^{3/4}} \]

[Out]

-1/8*arctan(2^(1/4)*x/(x^8+1)^(1/4))*2^(3/4)+1/8*arctan(2^(3/4)*x*(x^8+1)^(1/4)/(2^(1/2)*x^2-(x^8+1)^(1/2)))*2
^(1/4)-1/8*arctanh(2^(1/4)*x/(x^8+1)^(1/4))*2^(3/4)-1/8*arctanh(2*2^(1/4)*x*(x^8+1)^(1/4)/(2*x^2+2^(1/2)*(x^8+
1)^(1/2)))*2^(1/4)

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.14, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {440} \[ \int \frac {\left (1+x^8\right )^{3/4}}{-1+x^8} \, dx=-x \operatorname {AppellF1}\left (\frac {1}{8},1,-\frac {3}{4},\frac {9}{8},x^8,-x^8\right ) \]

[In]

Int[(1 + x^8)^(3/4)/(-1 + x^8),x]

[Out]

-(x*AppellF1[1/8, 1, -3/4, 9/8, x^8, -x^8])

Rule 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rubi steps \begin{align*} \text {integral}& = -x \operatorname {AppellF1}\left (\frac {1}{8},1,-\frac {3}{4},\frac {9}{8},x^8,-x^8\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.91 \[ \int \frac {\left (1+x^8\right )^{3/4}}{-1+x^8} \, dx=-\frac {\sqrt {2} \arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^8}}\right )-\arctan \left (\frac {2^{3/4} x \sqrt [4]{1+x^8}}{\sqrt {2} x^2-\sqrt {1+x^8}}\right )+\sqrt {2} \text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^8}}\right )+\text {arctanh}\left (\frac {2 \sqrt [4]{2} x \sqrt [4]{1+x^8}}{2 x^2+\sqrt {2} \sqrt {1+x^8}}\right )}{4\ 2^{3/4}} \]

[In]

Integrate[(1 + x^8)^(3/4)/(-1 + x^8),x]

[Out]

-1/4*(Sqrt[2]*ArcTan[(2^(1/4)*x)/(1 + x^8)^(1/4)] - ArcTan[(2^(3/4)*x*(1 + x^8)^(1/4))/(Sqrt[2]*x^2 - Sqrt[1 +
x^8])] + Sqrt[2]*ArcTanh[(2^(1/4)*x)/(1 + x^8)^(1/4)] + ArcTanh[(2*2^(1/4)*x*(1 + x^8)^(1/4))/(2*x^2 + Sqrt[2
]*Sqrt[1 + x^8])])/2^(3/4)

Maple [A] (verified)

Time = 24.62 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.09

method	result	size
pseudoelliptic	\(\frac {2^{\frac {1}{4}} \left (2 \arctan \left (\frac {\left (x^{8}+1\right )^{\frac {1}{4}} 2^{\frac {3}{4}}}{2 x}\right ) \sqrt {2}-\ln \left (\frac {2^{\frac {1}{4}} x +\left (x^{8}+1\right )^{\frac {1}{4}}}{-2^{\frac {1}{4}} x +\left (x^{8}+1\right )^{\frac {1}{4}}}\right ) \sqrt {2}+\ln \left (\frac {-\left (x^{8}+1\right )^{\frac {1}{4}} 2^{\frac {3}{4}} x +\sqrt {2}\, x^{2}+\sqrt {x^{8}+1}}{\left (x^{8}+1\right )^{\frac {1}{4}} 2^{\frac {3}{4}} x +\sqrt {2}\, x^{2}+\sqrt {x^{8}+1}}\right )+2 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{8}+1\right )^{\frac {1}{4}}+x}{x}\right )+2 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{8}+1\right )^{\frac {1}{4}}-x}{x}\right )\right )}{16}\)	\(165\)

[In]

int((x^8+1)^(3/4)/(x^8-1),x,method=_RETURNVERBOSE)

[Out]

1/16*2^(1/4)*(2*arctan(1/2*(x^8+1)^(1/4)/x*2^(3/4))*2^(1/2)-ln((2^(1/4)*x+(x^8+1)^(1/4))/(-2^(1/4)*x+(x^8+1)^(
1/4)))*2^(1/2)+ln((-(x^8+1)^(1/4)*2^(3/4)*x+2^(1/2)*x^2+(x^8+1)^(1/2))/((x^8+1)^(1/4)*2^(3/4)*x+2^(1/2)*x^2+(x
^8+1)^(1/2)))+2*arctan((2^(1/4)*(x^8+1)^(1/4)+x)/x)+2*arctan((2^(1/4)*(x^8+1)^(1/4)-x)/x))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 10.12 (sec) , antiderivative size = 603, normalized size of antiderivative = 3.97 \[ \int \frac {\left (1+x^8\right )^{3/4}}{-1+x^8} \, dx=-\frac {1}{32} \cdot 2^{\frac {3}{4}} \log \left (-\frac {4 \cdot 2^{\frac {1}{4}} {\left (x^{8} + 1\right )}^{\frac {1}{4}} x^{3} + 2 \cdot 2^{\frac {3}{4}} {\left (x^{8} + 1\right )}^{\frac {3}{4}} x + 4 \, \sqrt {x^{8} + 1} x^{2} + \sqrt {2} {\left (x^{8} + 2 \, x^{4} + 1\right )}}{x^{8} - 2 \, x^{4} + 1}\right ) + \frac {1}{32} \cdot 2^{\frac {3}{4}} \log \left (\frac {4 \cdot 2^{\frac {1}{4}} {\left (x^{8} + 1\right )}^{\frac {1}{4}} x^{3} + 2 \cdot 2^{\frac {3}{4}} {\left (x^{8} + 1\right )}^{\frac {3}{4}} x - 4 \, \sqrt {x^{8} + 1} x^{2} - \sqrt {2} {\left (x^{8} + 2 \, x^{4} + 1\right )}}{x^{8} - 2 \, x^{4} + 1}\right ) - \frac {1}{32} i \cdot 2^{\frac {3}{4}} \log \left (\frac {4 i \cdot 2^{\frac {1}{4}} {\left (x^{8} + 1\right )}^{\frac {1}{4}} x^{3} - 2 i \cdot 2^{\frac {3}{4}} {\left (x^{8} + 1\right )}^{\frac {3}{4}} x - 4 \, \sqrt {x^{8} + 1} x^{2} + \sqrt {2} {\left (x^{8} + 2 \, x^{4} + 1\right )}}{x^{8} - 2 \, x^{4} + 1}\right ) + \frac {1}{32} i \cdot 2^{\frac {3}{4}} \log \left (\frac {-4 i \cdot 2^{\frac {1}{4}} {\left (x^{8} + 1\right )}^{\frac {1}{4}} x^{3} + 2 i \cdot 2^{\frac {3}{4}} {\left (x^{8} + 1\right )}^{\frac {3}{4}} x - 4 \, \sqrt {x^{8} + 1} x^{2} + \sqrt {2} {\left (x^{8} + 2 \, x^{4} + 1\right )}}{x^{8} - 2 \, x^{4} + 1}\right ) + \left (\frac {1}{32} i - \frac {1}{32}\right ) \cdot 2^{\frac {1}{4}} \log \left (\frac {\left (2 i + 2\right ) \cdot 2^{\frac {3}{4}} {\left (x^{8} + 1\right )}^{\frac {1}{4}} x^{3} + 4 \, \sqrt {x^{8} + 1} x^{2} - \left (2 i - 2\right ) \cdot 2^{\frac {1}{4}} {\left (x^{8} + 1\right )}^{\frac {3}{4}} x + \sqrt {2} {\left (-i \, x^{8} + 2 i \, x^{4} - i\right )}}{x^{8} + 2 \, x^{4} + 1}\right ) - \left (\frac {1}{32} i + \frac {1}{32}\right ) \cdot 2^{\frac {1}{4}} \log \left (\frac {-\left (2 i - 2\right ) \cdot 2^{\frac {3}{4}} {\left (x^{8} + 1\right )}^{\frac {1}{4}} x^{3} + 4 \, \sqrt {x^{8} + 1} x^{2} + \left (2 i + 2\right ) \cdot 2^{\frac {1}{4}} {\left (x^{8} + 1\right )}^{\frac {3}{4}} x + \sqrt {2} {\left (i \, x^{8} - 2 i \, x^{4} + i\right )}}{x^{8} + 2 \, x^{4} + 1}\right ) + \left (\frac {1}{32} i + \frac {1}{32}\right ) \cdot 2^{\frac {1}{4}} \log \left (\frac {\left (2 i - 2\right ) \cdot 2^{\frac {3}{4}} {\left (x^{8} + 1\right )}^{\frac {1}{4}} x^{3} + 4 \, \sqrt {x^{8} + 1} x^{2} - \left (2 i + 2\right ) \cdot 2^{\frac {1}{4}} {\left (x^{8} + 1\right )}^{\frac {3}{4}} x + \sqrt {2} {\left (i \, x^{8} - 2 i \, x^{4} + i\right )}}{x^{8} + 2 \, x^{4} + 1}\right ) - \left (\frac {1}{32} i - \frac {1}{32}\right ) \cdot 2^{\frac {1}{4}} \log \left (\frac {-\left (2 i + 2\right ) \cdot 2^{\frac {3}{4}} {\left (x^{8} + 1\right )}^{\frac {1}{4}} x^{3} + 4 \, \sqrt {x^{8} + 1} x^{2} + \left (2 i - 2\right ) \cdot 2^{\frac {1}{4}} {\left (x^{8} + 1\right )}^{\frac {3}{4}} x + \sqrt {2} {\left (-i \, x^{8} + 2 i \, x^{4} - i\right )}}{x^{8} + 2 \, x^{4} + 1}\right ) \]

[In]

integrate((x^8+1)^(3/4)/(x^8-1),x, algorithm="fricas")

[Out]

-1/32*2^(3/4)*log(-(4*2^(1/4)*(x^8 + 1)^(1/4)*x^3 + 2*2^(3/4)*(x^8 + 1)^(3/4)*x + 4*sqrt(x^8 + 1)*x^2 + sqrt(2
)*(x^8 + 2*x^4 + 1))/(x^8 - 2*x^4 + 1)) + 1/32*2^(3/4)*log((4*2^(1/4)*(x^8 + 1)^(1/4)*x^3 + 2*2^(3/4)*(x^8 + 1
)^(3/4)*x - 4*sqrt(x^8 + 1)*x^2 - sqrt(2)*(x^8 + 2*x^4 + 1))/(x^8 - 2*x^4 + 1)) - 1/32*I*2^(3/4)*log((4*I*2^(1
/4)*(x^8 + 1)^(1/4)*x^3 - 2*I*2^(3/4)*(x^8 + 1)^(3/4)*x - 4*sqrt(x^8 + 1)*x^2 + sqrt(2)*(x^8 + 2*x^4 + 1))/(x^
8 - 2*x^4 + 1)) + 1/32*I*2^(3/4)*log((-4*I*2^(1/4)*(x^8 + 1)^(1/4)*x^3 + 2*I*2^(3/4)*(x^8 + 1)^(3/4)*x - 4*sqr
t(x^8 + 1)*x^2 + sqrt(2)*(x^8 + 2*x^4 + 1))/(x^8 - 2*x^4 + 1)) + (1/32*I - 1/32)*2^(1/4)*log(((2*I + 2)*2^(3/4
)*(x^8 + 1)^(1/4)*x^3 + 4*sqrt(x^8 + 1)*x^2 - (2*I - 2)*2^(1/4)*(x^8 + 1)^(3/4)*x + sqrt(2)*(-I*x^8 + 2*I*x^4
- I))/(x^8 + 2*x^4 + 1)) - (1/32*I + 1/32)*2^(1/4)*log((-(2*I - 2)*2^(3/4)*(x^8 + 1)^(1/4)*x^3 + 4*sqrt(x^8 +
1)*x^2 + (2*I + 2)*2^(1/4)*(x^8 + 1)^(3/4)*x + sqrt(2)*(I*x^8 - 2*I*x^4 + I))/(x^8 + 2*x^4 + 1)) + (1/32*I + 1
/32)*2^(1/4)*log(((2*I - 2)*2^(3/4)*(x^8 + 1)^(1/4)*x^3 + 4*sqrt(x^8 + 1)*x^2 - (2*I + 2)*2^(1/4)*(x^8 + 1)^(3
/4)*x + sqrt(2)*(I*x^8 - 2*I*x^4 + I))/(x^8 + 2*x^4 + 1)) - (1/32*I - 1/32)*2^(1/4)*log((-(2*I + 2)*2^(3/4)*(x
^8 + 1)^(1/4)*x^3 + 4*sqrt(x^8 + 1)*x^2 + (2*I - 2)*2^(1/4)*(x^8 + 1)^(3/4)*x + sqrt(2)*(-I*x^8 + 2*I*x^4 - I)
)/(x^8 + 2*x^4 + 1))

Sympy [F]

\[ \int \frac {\left (1+x^8\right )^{3/4}}{-1+x^8} \, dx=\int \frac {\left (x^{8} + 1\right )^{\frac {3}{4}}}{\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \left (x^{4} + 1\right )}\, dx \]

[In]

integrate((x**8+1)**(3/4)/(x**8-1),x)

[Out]

Integral((x**8 + 1)**(3/4)/((x - 1)*(x + 1)*(x**2 + 1)*(x**4 + 1)), x)

Maxima [F]

\[ \int \frac {\left (1+x^8\right )^{3/4}}{-1+x^8} \, dx=\int { \frac {{\left (x^{8} + 1\right )}^{\frac {3}{4}}}{x^{8} - 1} \,d x } \]

[In]

integrate((x^8+1)^(3/4)/(x^8-1),x, algorithm="maxima")

[Out]

integrate((x^8 + 1)^(3/4)/(x^8 - 1), x)

Giac [F]

\[ \int \frac {\left (1+x^8\right )^{3/4}}{-1+x^8} \, dx=\int { \frac {{\left (x^{8} + 1\right )}^{\frac {3}{4}}}{x^{8} - 1} \,d x } \]

[In]

integrate((x^8+1)^(3/4)/(x^8-1),x, algorithm="giac")

[Out]

integrate((x^8 + 1)^(3/4)/(x^8 - 1), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (1+x^8\right )^{3/4}}{-1+x^8} \, dx=\int \frac {{\left (x^8+1\right )}^{3/4}}{x^8-1} \,d x \]

[In]

int((x^8 + 1)^(3/4)/(x^8 - 1),x)

[Out]

int((x^8 + 1)^(3/4)/(x^8 - 1), x)